求证x/(y+z-x)+y/(x+z-y)+z/(x+y-z)≥3

Question

求证x/(y+z-x)+y/(x+z-y)+z/(x+y-z)≥3

Answer 1

x/(y+z-x)+y/(x+z-y)+z/(x+y-z)≥3
令y+z-x=a,x+z-y=b,x+y-z=c
z=a+b/2 x=b+c/2 y=a+c/2
原式化为：
证明b+c/2a + a+c/2b+a+b/2c≥3
即证明b/a+c/a+a/b+c/b+a/c+b/c≥6
b/a+a/b+c/a+a/c+b/c+c/b≥6
好像给的条件不够啊,如果满足左边各项均大于0,下面接着用不等式定理好了!
b/a+a/b+c/a+a/c+b/c+c/b≥2+2+2＝6
条件不够,举个反例子：x=0 y=0 z=1
代进去为－1 不满足